In an IMO SL problem for 2020 number theory problem 6, there is a part of the solution which I do not get which states the following:
Let $m>5$ be large positive integer and also let $n=p_1p_2...p_{\pi(m)}$ be the product of all primes not exceeding $m$. Then $\phi (d(n))=\phi (2^{\pi(m)})=2^{\pi(m)-1}$. Consider the following:
$$\phi(n)=\prod_{k=1}^{\pi(m)}(p_k-1)=\prod_{s=1}^{\pi(m/2)}q_s^{\alpha_s}$$
where $q_1q_2...q_{\pi(m/2)}$ are primes not exceeding $m/2$. Note that every term $p_k-1$ contributes at most one prime $q_s>\sqrt{m}$ in the product $\prod_{s=1}^{\pi(m/2)}q_s^{\alpha_s}$. Hence we have:
$$\sum_{s: \space q_s>\sqrt{m}}\alpha_s \le \pi(m) \rightarrow \sum_{s: \space q_s>\sqrt{m}}(1+\alpha_s) \le \pi(m)+\pi(m/2)$$
My question:
How do we know that the prime cannot exceed $\sqrt{m}$ ?
How can we just state that $\phi(n)=\prod_{k=1}^{\pi(m)}(p_k-1)=\prod_{s=1}^{\pi(m/2)}q_s^{\alpha_s}$, I do not understand the motivation behind the second product of primes $q_{s}$ and its bound on the product of $\pi(m/2)$.
In the last inequality how does $\sum_{s: \space q_s>\sqrt{m}}\alpha_s \le \pi(m)$ lead to $\sum_{s: \space q_s>\sqrt{m}}(1+\alpha_s) \le \pi(m)+\pi(m/2)$ ?
I want to understand the remaining of the solution but am stuck on these parts, could anyone kindly explain how those hold true ?